If $f:X\to Y$ and you have an ultrafilter $G$ on $X$, then it induces an ultrafilter $U$ on $Y$ by defining $A\in U\iff f^{-1}A\in G$. The ultrafilter $U$ is said to be Rudin-Kiesler below $G$, and this ordering on ultrafilters is intensely studied in large cardinal set theory. It follows that $f[G]\subset U$, and so this may be the answer you seek.

But in general, I am not sure what you mean by the *image* filter, when your maps go in that direction. Perhaps there is a typo?

Perhaps you meant to ask the following question: you have a map $f:X\to Y$ and a filter $F$ on $X$ (not $Y$), and the image $f[F]\subset G$ for some ultrafilter $G$ on $Y$. The question is whether there is an ultrafilter $U$ on $X$ with $f[U]\subset G$.

In this case, let $U$ be any ultrafilter containing $F$ and all $f^{-1}A$ for $A\in G$. There is such an ultrafilter (as Kevin also explains in his answer) since if $B\in F$ and $A\in G$, then $f[B]\cap A\in G$ and so $B\cap f^{-1}A$ is nonempty (and so the collection forms a filter base). For any such $U$, we have $f[U]\subset G$. If $f$ is onto $Y$, then we get actually $f[U]=G$.

Note, if $f$ is not onto $Y$, then there is no possiblity that $f[U]=G$, if this should mean $G$ consists precisely of $f[A]$ for $A\in U$, since all such $f[A]$ are subsets of the range of $f$. (So in this technical sense, the answer to your question is negative.) But on the positive side, $G$ is the image of $U$ in the sense that it is the filter gnerated by the sets in $f[U]$.